Barycentric Calculus

Date: 01/06/99 at 09:33:07
From: A.H.Noble
Subject: Barycentric Calculus

Please give me infomation on the workings of Barycentric Calculus. I 
first came across the subject in David Wells' book, _You Are a 
Mathematician_, which gives a tantalizing intoduction, and I would like 
to know more.

Date: 01/08/99 at 08:07:22
From: Doctor Floor
Subject: Re: Barycentric Calculus

Hello!

Thank you for sending your question to Dr. Math.

Barycentric coordinates (or "barycentrics") are coordinates with 
respect to a reference triangle ABC.

The original interpretation of the point P given by barycentric 
coordinates p:q:r is that P is the center of mass ("barycenter") of 
weights p, q and r when these are placed on the vertices A, B and C.

Another interpretation is that the ratio of the areas BCP:ACP:ABP is 
p:q:r.

These interpretations are equivalent. It is also to be understood that 
for any nonzero real t p:q:r and tp:tq:tr represent the same point. 
For this reason barycentric coordinates are said to be "homogeneous." 

Another example of homogeneous coordinates with respect to a triangle 
is the notion of trilinear coordinates. Trilinear coordinates are 
given by the ratio p:q:r of the distances to the sidelines of the 
reference triangle. A short introduction on trilinear coordinates 
(trilinears) can be found at:

   http://faculty.evansville.edu/ck6/tcenters/trilin.html   

Barycentrics have a lot of advantages over trilinear. They have been 
discussed in the geometry.research newsgroup. You can find the most 
interesting letter from the discussion, for the largest part consisting 
of a quote by Professor John Conway, at:

 http://mathforum.org/kb/message.jspa?messageID=1091956   

(Select 30 Jun 1998, "Conway on Trilinear vs Barycentric coordinates, 
by steve sigur.")

When a point has trilinears t:u:v then its barycentrics are at:bu:cv, 
where a,b,c are the sidelengths of the reference triangle.

The barycentric coordinates of the vertices of the reference triangle 
are simple:

   A 1:0:0
   B 0:1:0
   C 0:0:1

Three points given by a1:b1:c1, a2:b2:c2 and a3:b3:c3 are collinear if 
and only if the determinant:

  | a1 b1 c1 |
  | a2 b2 c2 |
  | a3 b3 c3 |

equals zero. This means that the line through the given points a1:b1:c1 
and a2:b2:c2 is found by points x:y:z that satisfy:

  | x  y  z  |
  | a1 b1 c1 | = 0        [1]
  | a2 b2 c2 |

This can also be written as:

  (b1c2-b2c1) x + (c1a2-c2a1) y + (a1b2-a2b1) z = 0         [2]

From this we find the general appearance of a line formed by points 
x:y:z that satisfy:

  ax + by + cz = 0 for some a,b,c that are not all zero.

There is a lot of similarity in computations with lines and with 
points:

Three lines with coefficients  a1,b1,c1  a2,b2,c2  and  a3,b3,c3  are 
collinear if and only if the determinant:

  | a1 b1 c1 |
  | a2 b2 c2 | = 0    (Cf. [1])
  | a3 b3 c3 |

The point of intersection of two lines a1x+b1y+c1z = 0 and 
a2x+b2y+c3z = 0 is given by:

  b1c2-b2c1 : c1a2-c2a1 : a1b2-a2b1   (Cf. [2])

Barycentrics t:u:v are called "normalized" if t+u+v = 1. Normalized 
barycentrics have a lot of interesting properties, especially when one 
wants to calculate actual distances, areas, etc.

Among the nice features of normalized barycentrics is also that one 
can use them to calculate triangle centers. For instance, the 
normalized barycentrics of the centroid are always given by 
1/3:1/3:1/3.

When we know the coordinates (barycentric, trilinear or even cartesian) 
of a triangle we can use the barycentrics to calculate the actual 
point.

Let A(1,5), B(3,8), and C(5,-1) form the reference triangle. Then the 
point P with normalized barycentrics t:u:v is given by:

   P(t*1 + u*3 + v*5, t*5 + u*8 + v*(-1))

So the centroid of ABC is P(3,4).

Barycentrics of triangle centers are usually given as functions of 
sidelengths a, b and c and/or angles A, B and C.

As an example, the incenter of a triangle ABC always has barycentrics 
a:b:c. Triangle centers can be found - given in trilinears - at:

   http://faculty.evansville.edu/ck6/tcenters/index.html   

I hope this has helped. If you have a math question again, please send 
it to Dr. Math!

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/